Localized Current Distribution

Consider the magnetic field generated by a current distribution that is localized in some relatively small region of space centered on the origin. From Equation (621), we have

$${\bf A}({\bf r}) = \frac{\mu_0}{4\pi}\int\frac{{\bf j}({\bf r}')}{\vert{\bf r}-{\bf r}'\vert}\,dV'.$$ (661)

Assuming that $r\gg r'$ , so that our observation point lies well outside the distribution, we can write

$$\frac{1}{\vert{\bf r}-{\bf r}'\vert}= \frac{1}{\vert{\bf r}\vert}+\frac{{\bf r}\cdot{\bf r}'}{\vert{\bf r}\vert^{\,3}}+\cdots.$$ (662)

Thus, the $i$ th Cartesian component of the vector potential has the expansion

$$A_i({\bf r})= \frac{\mu_0}{4\pi}\,\frac{1}{\vert{\bf r}\vert}\int... ...{{\bf r}}{\vert{\bf r}\vert^{\,3}}\cdot\int j_i({\bf r}')\,{\bf r}'\,dV'+\cdots$$ (663)

Consider the integral

$$K = \int (f\,{\bf j}\cdot \nabla' g+g\,{\bf j}\cdot\nabla' f)\,dV',$$ (664)

where ${\bf j}({\bf r}')$ is a divergence-free [see Equation (618)] localized current distribution, and $f({\bf r}')$ and $g({\bf r}')$ are two well-behaved functions. Integrating the first term by parts, making use of the fact that ${\bf j}'({\bf r}')\rightarrow {\bf0}$ as $\vert{\bf r}'\vert\rightarrow\infty$ (because the current distribution is localized), we obtain

$$K=\int\left[-g\,\nabla'\cdot(f\,{\bf j})+g\,{\bf j}\cdot\nabla' f\right]dV'$$ (665)

Hence,

$$K = \int\left[-g\,{\bf j}\cdot\nabla' f -g\,f\,\nabla'\cdot{\bf j}+g\,{\bf j}\cdot\nabla' f\right]dV'=0,$$ (666)

because $\nabla'\cdot{\bf j}= 0$ . Thus, we have proved that

$$\int (f\,{\bf j}\cdot \nabla' g+g\,{\bf j}\cdot\nabla' f)\,dV'=0.$$ (667)

Let $f=1$ and $g=x_i'$ (where $x_i'$ is the $i$ th component of ${\bf r}'$ ). It immediately follows from Equation (668) that

$$\int j_i({\bf r}')\,dV'= 0.$$ (668)

Likewise, if $f=x_i'$ and $g=x_j'$ then Equation (668) implies that

$$\int \left(x_i'\,j_j + x_j'\,j_i\right)\,dV' = 0.$$ (669)

According to Equations (664) and (669),

$$A_i({\bf r})= \frac{\mu_0}{4\pi}\,\frac{{\bf r}}{\vert{\bf r}\vert^{\,3}}\cdot\int j_i({\bf r}')\,{\bf r}'\,dV'+\cdots.$$ (670)

Now,

$${\bf r}\cdot\int j_i({\bf r}')\,{\bf r}'\,dV' = x_j\int x'_j\,j_i\,dV'= -\frac{1}{2}x_j\int(x_i'\,j_j-x_j'\,j_i)\,dV',$$ (671)

where use has been made of Equation (670), as well as the Einstein summation convention. Thus,

$${\bf r}\cdot\int j_i\,{\bf r}'\,dV' = -\frac{1}{2}\int \left[({\b... ... = -\frac{1}{2}\left[{\bf r}\times \int ({\bf r}'\times {\bf j})\,dV'\right]_i.$$ (672)

Hence, we obtain

$${\bf A}({\bf r})=-\frac{\mu_0}{8\pi}\,\frac{{\bf r}}{\vert{\bf r}\vert^{\,3}}\times \int ({\bf r}'\times {\bf j})\,dV'+\cdots.$$ (673)

It is conventional to define the magnetization, or magnetic moment density, as

$${\bf M}({\bf r}) = \frac{1}{2}\,{\bf r}\times {\bf j}({\bf r}).$$ (674)

The integral of this quantity is known as the magnetic moment:

$${\bf m} = \frac{1}{2}\int{\bf r}'\times {\bf j}'({\bf r}')\,dV'.$$ (675)

It immediately follows from Equation (674) that the vector potential a long way from a localized current distribution takes the form

$${\bf A}({\bf r}) = \frac{\mu_0}{4\pi}\,\frac{{\bf m}\times {\bf r}}{r^{\,3}}.$$ (676)

The corresponding magnetic field is

$${\bf B}({\bf r}) = \nabla\times {\bf A} = \frac{\mu_0}{4\pi}\left[\frac{3\,({\bf m}\cdot{\bf r})\,{\bf r} -r^{\,2}\,{\bf m}}{r^{\,5}}\right].$$ (677)

Thus, we have demonstrated that the magnetic field far from any localized current distribution takes the form of a magnetic dipole field whose moment is given by the integral (676).

Consider a localized current distribution that consists of a closed planar loop carrying the current $I$ . If $d{\bf r}$ is a line element of the loop then Equation (676) reduces to

$${\bf m} = I\oint\frac{1}{2}\,{\bf r}\times d{\bf r}.$$ (678)

However, $(1/2)\,{\bf r}\times d{\bf r}=d{\bf A}$ , where $d{\bf A}$ is a triangular element of vector area defined by the two ends of $d{\bf r}$ and the origin. Thus, the loop integral gives the total vector area, ${\bf A}$ , of the loop. It follows that

$${\bf m} = I\,A\,{\bf n},$$ (679)

where ${\bf n}$ is a unit normal to the loop in the sense determined by the right-hand circulation rule (with the current determining the sense of circulation). Of course, Equation (680) is identical to Equation (659).